find the derivative of cos root x wrt x from first principle - Maths - Continuity and Differentiability

Question

find the derivative of cos root x wrt x from first
principle??

Lalit Mehra · Accepted Answer

Given, f (x) = $\cos \sqrt{x}$

$\frac{d y}{d x} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} (First principle) ∴ \frac{d y}{d x} = \lim_{h \to 0} \frac{\cos \sqrt{x + h} - \cos \sqrt{x}}{h} \Rightarrow \frac{d y}{d x} = \lim_{h \to 0} \frac{- 2 \sin (\frac{\sqrt{x + h} + \sqrt{x}}{2}) \sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{h} \Rightarrow \frac{d y}{d x} = - 2 \lim_{h \to 0} \sin (\frac{\sqrt{x + h} + \sqrt{x}}{2}) \times \lim_{h \to 0} \frac{\sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{(\frac{\sqrt{x + h} - \sqrt{x}}{2})} \times \frac{(\sqrt{x + h} - \sqrt{x})}{2 h}$

$\Rightarrow \frac{d y}{d x} = - \sin (\frac{\sqrt{x + 0} + \sqrt{x}}{2}) \times \lim_{h \to 0} \frac{\sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{(\frac{\sqrt{x + h} - \sqrt{x}}{2})} \times \lim_{h \to 0} \frac{(\sqrt{x + h} - \sqrt{x})}{h} \times \frac{(\sqrt{x + h} + \sqrt{x})}{\sqrt{x + h} + \sqrt{x}} \Rightarrow \frac{d y}{d x} = - \sin \sqrt{x} \times 1 \times \lim_{h \to 0} \frac{(x + h - x)}{h (\sqrt{x + h} + \sqrt{x})} \Rightarrow \frac{d y}{d x} = - \sin \sqrt{x} \times 1 \times \frac{1}{\lim_{h \to 0} (\sqrt{x + h} + \sqrt{x})} \Rightarrow \frac{d y}{d x} = - \sin \sqrt{x} \times 1 \times \frac{1}{(\sqrt{x} + \sqrt{x})} \Rightarrow \frac{d y}{d x} = - \frac{\sin \sqrt{x}}{2 \sqrt{x}}$

find the derivative of cos root x wrt x from first principle??